Exploring the Accumulation Points of $1/n$

[Dustinsfl]

All numbers of the form $1/n$, $(n = 1,2,3,\ldots)$.$1/n = (0,1)$
The accumulation points are $x\in [0,1]$.
This set is open.

 

[Sudharaka]

All numbers of the form $1/n$, $(n = 1,2,3,\ldots)$.$1/n = (0,1)$
The accumulation points are $x\in [0,1]$.
This set is open.

Hi dwsmith, :)

So the set under consideration is, \(S=\{\frac{1}{n}:n\in\mathbb{Z}^+\}\). The definition of a limit point (accumulation point) taking the set of real numbers as the reference space is given >>here<<. If you go by that definition you can prove that the only accumulation point of \(S\) is zero.

Kind Regards,
Sudharaka.

 

[Amer]

All numbers of the form $1/n$, $(n = 1,2,3,\ldots)$.$1/n = (0,1)$
The accumulation points are $x\in [0,1]$.
This set is open.

$1/n = (0,1)$ this is not true

 

[Dustinsfl]

Hi dwsmith, :)

So the set under consideration is, \(S=\{\frac{1}{n}:n\in\mathbb{Z}^+\}\). The definition of a limit point (accumulation point) taking the set of real numbers as the reference space is given >>here<<. If you go by that definition you can prove that the only accumulation point of \(S\) is zero.

Kind Regards,
Sudharaka.

As with the complex example, why aren't the points of $1/n$ accumulation points as well?

 

[Sudharaka]

As with the complex example, why aren't the points of $1/n$ accumulation points as well?

Let me try to visualize things for you. The neighboring points of \(\frac{1}{n}\) are \(\frac{1}{n+1}\) and \(\frac{1}{n-1}\). You can take a neibourhood around \(\frac{1}{n}\) such that both of these points are not included. But to be a limit point every neighborhood of \(\frac{1}{n}\) should contain at least one point (in \(S\)) distinct from \(\frac{1}{n}\), which is not the case here. Does this clarify things for you?

Kind Regards,
Sudharaka.